question_answer

A particle is released from a height 5. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

A)  \[\frac{S}{4},\frac{3gS}{2}\]                       

B)  \[\frac{S}{4},\frac{\sqrt{3gS}}{2}\]         

C)         \[\frac{S}{2},\frac{\sqrt{3gS}}{2}\]         

D)         \[\frac{S}{2},\frac{\sqrt{3gS}}{2}\] 

E)  \[\frac{S}{3},\frac{\sqrt{3gS}}{2}\]

Correct Answer: D

Solution :

We can realise the situation as shown. Let at point C distance\[x\]from highest point A, the particles kinetic energy is three times its potential energy.                Velocity at C,                     \[{{v}^{2}}=0+2gx\] or          \[{{v}^{2}}=2gx\]                                            ...(i) Potential energy at C \[=mg(S-x)\]                                      ...(ii) At point C, Kinetic energy\[=3\times \] potential energy ie,           \[\frac{1}{2}m\times 2gx=3\times mg(S-x)\] or          \[x=3S-3x\] or          \[4x=3S\] or           \[S=\frac{4}{3}x\] or          \[x=\frac{3}{4}S\] Therefore, from Eq. (i) \[{{v}^{2}}=2g\times \frac{3}{4}S\] Or           \[{{v}^{2}}=\frac{3}{2}gS\]or\[V=\sqrt{\frac{3}{2}gS}\] Height of the particle from the ground \[=S-x\]                 \[=S-\frac{3}{4}S=\frac{S}{4}\]